HDU 1905 Pseudoprime numbers (快速冪求餘)
Description
Fermat’s theorem states that for any prime number p and for any integer a > 1, a^p == a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, known as base-a pseudoprimes, have this property for some a. (And some, known as Carmichael Numbers, are base-a pseudoprimes for all a.)
Given 2 < p ≤ 1,000,000,000 and 1 < a < p, determine whether or not p is a base-a pseudoprime.
Input
Input contains several test cases followed by a line containing “0 0”. Each test case consists of a line containing p and a.
Output
For each test case, output “yes” if p is a base-a pseudoprime; otherwise output “no”.
Sample Input
3 2
10 3
341 2
341 3
1105 2
1105 3
0 0
Sample Output
no
no
yes
no
yes
yes
快速冪求餘,原理我沒看懂,只能當模板用了~
#include <iostream>
using namespace std;
bool isprime(long long a) {
for(long long i = 2; i * i <= a; i++) {
if (a % i == 0) return false;
}
return true;
}
long long qmod(long long a, long long r, long long m) {
long long res = 1;
while (r) {
if (r & 1)
res = res * a % m;
a = a * a % m;
r >>= 1;
}
return res;
}
int main() {
long long p, a;
while (scanf("%I64d%I64d", &p, &a) && p && a) {
if (!isprime(p) && qmod(a, p, p) == a)
printf("yes\n");
else
printf("no\n");
}
return 0;
}